8.4 Conditional Probability
8.4 Conditional Probability
Learning Objectives
By the end of this section, you will be able to:
- Recognize situations that involve conditional probability.
- Compute conditional probabilities directly from a reduced sample space and from the formula \(P(E \mid F) = P(E \cap F)/P(F)\).
8.4.1 What "Given That" Means
Suppose a friend asks for the probability that it will snow today. The honest answer depends on where you are.
- In Boston in winter, the probability is fairly high.
- In Cupertino in summer, the probability is essentially zero.
Letting \(A\) = "it snows today," \(B\) = "I'm in Boston in winter," and \(C\) = "I'm in Cupertino in summer," we cannot give a single \(P(A)\); we must specify the condition. We write \(P(A \mid B)\) for the probability of \(A\) given \(B\), and \(P(A \mid C)\) for the probability of \(A\) given \(C\).
The vertical bar \(\mid\) is read "given that" or "if we know that". The event of interest sits to the left of the bar; the condition sits to the right.
Context Pause: Why conditional probability shows up everywhere
Almost every real probability question is conditional. Risk of accident given a 16-year-old driver. Likelihood of a defect given the part came from supplier B. Probability of disease given a positive test. The unconditional version is mathematically simpler, but the conditional version is what actually informs decisions, because we usually know something about the situation.
Insight Note: Conditioning shrinks the sample space
Knowing that condition \(F\) holds removes every outcome not in \(F\) from consideration. The new "denominator" is \(n(F)\), and the new "numerator" is the number of outcomes that lie in both the event of interest and \(F\). Drawing the picture (a Venn diagram with the relevant region shaded) makes the formula self-evident.
8.4.2 Computing Conditional Probability from a Reduced Sample Space
When the sample space is small, you can list it, condition on \(F\), and just count.
Example 8.4.1
A family has three children. Find the conditional probability of having two boys and a girl, given that the first born is a boy.
Step 1 — Full sample space: \[S = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}, \quad n(S) = 8.\] Step 2 — Condition on "first born is a boy". That cuts the sample space to outcomes starting with \(B\): \[F = \{BBB, BBG, BGB, BGG\}, \quad n(F) = 4.\] Step 3 — Count outcomes in \(F\) that also have two boys and one girl: \(BBG\) and \(BGB\). Step 4 — Compute: \[P(E \mid F) = \frac{2}{4} = \frac{1}{2}.\] Answer: \(1/2\).Solution
Example 8.4.2
A six-sided die is rolled once. Let \(E = \{\text{the result is even}\}\) and \(T = \{\text{the result is greater than 3}\}\). Find:
a) \(P(E)\).
b) \(P(E \mid T)\).
Sample space: \(S = \{1, 2, 3, 4, 5, 6\}\). \(E = \{2, 4, 6\}\), \(T = \{4, 5, 6\}\). Part a — Unconditional. \(P(E) = 3/6 = 1/2\). Part b — Conditional. Given \(T\) holds, the only possible outcomes are \(\{4, 5, 6\}\). Of those, the even ones are \(\{4, 6\}\). Therefore \[P(E \mid T) = \frac{2}{3}.\]Solution
Example 8.4.3
A fair coin is tossed twice. Find:
a) \(P(\text{two heads})\).
b) \(P(\text{two heads} \mid \text{at least one head})\).
Sample space: \(S = \{HH, HT, TH, TT\}\). Let \(E = \{\text{two heads}\} = \{HH\}\) and \(F = \{\text{at least one head}\} = \{HH, HT, TH\}\). Part a — Unconditional. \(P(E) = 1/4\). Part b — Conditional. Given \(F\), the reduced sample space is \(\{HH, HT, TH\}\) with \(n(F) = 3\). Of those, only \(HH\) gives two heads: \[P(E \mid F) = \frac{1}{3}.\]Solution
Try It Now 8.4.1
A single die is rolled. Find \(P(\text{the result is a 3} \mid \text{the result is odd})\).
The conditioning event "odd" is \(\{1, 3, 5\}\), with \(n = 3\). The event "result is 3" inside that reduced space is \(\{3\}\). Therefore \[P = \frac{1}{3}.\] Answer: \(1/3\).Solution
8.4.3 The Conditional Probability Formula
When the sample space is too large to enumerate, we use a formula. Suppose the experiment has \(n\) equally likely outcomes; suppose \(F\) has \(m\) outcomes and \(E \cap F\) has \(c\) outcomes. Conditioning on \(F\) restricts attention to those \(m\) outcomes; among them, \(c\) are favorable to \(E\). So
\[P(E \mid F) = \frac{c}{m} = \frac{c/n}{m/n} = \frac{P(E \cap F)}{P(F)}.\]
Conditional Probability Formula
For events \(E\) and \(F\) with \(P(F) > 0\),
\[P(E \mid F) \;=\; \frac{P(E \cap F)}{P(F)}.\]
Equivalently, the multiplication rule:
\[P(E \cap F) \;=\; P(F)\, P(E \mid F).\]
Example 8.4.4
Suppose \(P(A) = 0.3\), \(P(B) = 0.4\), and \(P(A \cap B) = 0.12\). Find \(P(A \mid B)\) and \(P(B \mid A)\).
\[P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = 0.30.\] \[P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.3} = 0.40.\] Answer: \(P(A \mid B) = 0.30\), \(P(B \mid A) = 0.40\).Solution
Try It Now 8.4.2
If \(P(F) = 0.4\) and \(P(E \mid F) = 0.3\), find \(P(E \cap F)\).
Use the multiplication rule: \[P(E \cap F) = P(F)\, P(E \mid F) = 0.4 \times 0.3 = 0.12.\] Answer: \(0.12\).Solution
8.4.4 Conditional Probability from Two-Way Tables
A two-way table makes conditional probabilities very natural: condition on a row (or column) and read the relative frequency along that row (or column).
Example 8.4.5
The table below shows the distribution of the 100 U.S. senators of the 114th Congress (January 2015) by political party and gender.
| | Male (M) | Female (F) | Total | |--------------|---------:|-----------:|------:| | Democrats (D)| 30 | 14 | 44 | | Republicans (R)| 48 | 6 | 54 | | Other (T) | 2 | 0 | 2 | | Total | 80 | 20 | 100 |
Find:
a) \(P(M \mid D)\).
b) \(P(D \mid M)\).
c) \(P(F \mid R)\).
d) \(P(R \mid F)\).
Part a — Condition on Democrats. Among the 44 Democrats, 30 are male: \[P(M \mid D) = \frac{30}{44} = \frac{15}{22} \approx 0.682.\] Part b — Condition on Males. Among the 80 males, 30 are Democrats: \[P(D \mid M) = \frac{30}{80} = \frac{3}{8} = 0.375.\] Part c — Condition on Republicans. Among the 54 Republicans, 6 are female: \[P(F \mid R) = \frac{6}{54} = \frac{1}{9} \approx 0.111.\] Part d — Condition on Females. Among the 20 females, 6 are Republicans: \[P(R \mid F) = \frac{6}{20} = \frac{3}{10} = 0.30.\]Solution
Insight Note: \(P(A \mid B)\) and \(P(B \mid A)\) are not the same
Example 8.4.5 made this explicit: \(P(M \mid D) \approx 0.682\), but \(P(D \mid M) = 0.375\). The two ratios share a numerator (the count of male Democrats) but use different denominators (Democrats vs. males). Confusing the two — sometimes called the "prosecutor's fallacy" — is one of the most common probability errors.
Try It Now 8.4.3
Using the same Senate table, find \(P(M \mid F)\).
Among the 20 females, none are male: \[P(M \mid F) = \frac{0}{20} = 0.\] Answer: \(0\) (Senators in this table are exclusively male or female; the events are mutually exclusive).Solution
Problem Set 8.4
For problems 8.4.1–8.4.4, use the conditional probability formula \(P(A \mid B) = P(A \cap B)/P(B)\).
8.4.1 A card is drawn from a deck. Find \(P(\text{a queen} \mid \text{a face card})\).
8.4.2 A card is drawn from a deck. Find \(P(\text{a queen} \mid \text{a club})\).
8.4.3 A die is rolled. Find the conditional probability that it shows a three given that an odd number has shown.
8.4.4 If \(P(A) = 0.3\), \(P(B) = 0.4\), and \(P(A \cap B) = 0.12\), find:
a) \(P(A \mid B)\)
b) \(P(B \mid A)\)
For problems 8.4.5–8.4.8, use the Senate table:
| | Male (M) | Female (F) | Total | |--------------|---------:|-----------:|------:| | Democrats (D)| 30 | 14 | 44 | | Republicans (R)| 48 | 6 | 54 | | Other (T) | 2 | 0 | 2 | | Total | 80 | 20 | 100 |
8.4.5 \(P(M \mid D)\)
8.4.6 \(P(D \mid M)\)
8.4.7 \(P(F \mid R)\)
8.4.8 \(P(R \mid F)\)
Solve each conditional probability problem.
8.4.9 At a college, 20% of the students take Finite Math, 30% take History, and 5% take both Finite Math and History. If a student is chosen at random, find:
a) \(P(\text{Finite Math} \mid \text{History})\).
b) \(P(\text{History} \mid \text{Finite Math})\).
8.4.10 At a college, 60% of students pass Accounting, 70% pass English, and 30% pass both. If a student is selected at random, find:
a) \(P(\text{Accounting} \mid \text{English})\).
b) \(P(\text{English} \mid \text{Accounting})\).
8.4.11 If \(P(F) = 0.4\) and \(P(E \mid F) = 0.3\), find \(P(E \cap F)\).
8.4.12 \(P(E) = 0.3\), \(P(F) = 0.3\), and \(E\) and \(F\) are mutually exclusive. Find \(P(E \mid F)\).
8.4.13 If \(P(E) = 0.6\) and \(P(E \cap F) = 0.24\), find \(P(F \mid E)\).
8.4.14 If \(P(E \cap F) = 0.04\) and \(P(E \mid F) = 0.1\), find \(P(F)\).
At a college, 72% of courses have final exams and 46% of courses require research papers. 32% of courses have both a research paper and a final exam. Let \(F\) be the event that a course has a final exam and \(R\) be the event that a course requires a research paper.
8.4.15 Find \(P(\text{course has a final exam} \mid \text{course has a research paper})\).
8.4.16 Find \(P(\text{course has a research paper} \mid \text{course has a final exam})\).
Consider a family of three children. Find the following conditional probabilities.
8.4.17 \(P(\text{two boys} \mid \text{first born is a boy})\)
8.4.18 \(P(\text{all girls} \mid \text{at least one girl is born})\)
8.4.19 \(P(\text{children of both sexes} \mid \text{first born is a boy})\)
8.4.20 \(P(\text{all boys} \mid \text{children of both sexes})\)
For problems 8.4.21–8.4.26, use the table below — the highest attained educational status for a sample of U.S. residents age 25+ (data: census.gov, Table 1-01, 2010).
| | Did not Complete HS (D) | HS Grad (H) | Some College (C) | Associate (A) | Bachelor (B) | Graduate (G) | Total | |--------|-----------------------:|------------:|-----------------:|--------------:|-------------:|-------------:|------:| | 25–44 (R) | 95 | 228 | 143 | 81 | 188 | 61 | 796 | | 45–64 (S) | 83 | 256 | 136 | 80 | 150 | 67 | 772 | | 65+ (T) | 96 | 191 | 84 | 36 | 80 | 41 | 528 | | Total | 274 | 675 | 363 | 197 | 418 | 169 | 2096 |
8.4.21 \(P(C \mid T)\)
8.4.22 \(P(S \mid A)\)
8.4.23 \(P(C \cap T)\)
8.4.24 \(P(R \mid B)\)
8.4.25 \(P(B \mid R)\)
8.4.26 \(P(G \mid S)\)
Step 1 — Apply the formula: \[P(\text{queen} \mid \text{face card}) = \frac{P(\text{queen} \cap \text{face card})}{P(\text{face card})}.\] Step 2 — Plug in: every queen is a face card, so \(P(\text{queen} \cap \text{face}) = 4/52\). \(P(\text{face card}) = 12/52\). \[P = \frac{4/52}{12/52} = \frac{4}{12} = \frac{1}{3}.\] Answer: \(1/3\).Problem 8.4.1 Solution
Step 1 — Apply the formula: \[P(\text{queen} \mid \text{club}) = \frac{P(\text{queen} \cap \text{club})}{P(\text{club})}.\] Step 2 — Compute. Only the Queen of Clubs is both a queen and a club, so \(P(\text{queen} \cap \text{club}) = 1/52\). \(P(\text{club}) = 13/52\). \[P = \frac{1/52}{13/52} = \frac{1}{13}.\] Answer: \(1/13\).Problem 8.4.2 Solution
Step 1 — Reduce the sample space to "odd": \(\{1, 3, 5\}\) with 3 outcomes. Step 2 — Count favorable: the result equals 3 in 1 of those. Step 3 — Compute: \[P(3 \mid \text{odd}) = \frac{1}{3}.\] Answer: \(1/3\).Problem 8.4.3 Solution
Part a: \[P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = 0.3.\] Part b: \[P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.3} = 0.4.\] Answer: a) \(0.3\); b) \(0.4\).Problem 8.4.4 Solution
Step 1 — Condition on Democrats (44 of them). 30 are male: \[P(M \mid D) = \frac{30}{44} = \frac{15}{22}.\] Answer: \(15/22 \approx 0.682\).Problem 8.4.5 Solution
Step 1 — Condition on males (80 of them). 30 are Democrats: \[P(D \mid M) = \frac{30}{80} = \frac{3}{8} = 0.375.\] Answer: \(3/8\).Problem 8.4.6 Solution
Step 1 — Condition on Republicans (54 of them). 6 are female: \[P(F \mid R) = \frac{6}{54} = \frac{1}{9}.\] Answer: \(1/9 \approx 0.111\).Problem 8.4.7 Solution
Step 1 — Condition on females (20 of them). 6 are Republicans: \[P(R \mid F) = \frac{6}{20} = \frac{3}{10} = 0.30.\] Answer: \(3/10\).Problem 8.4.8 Solution
Let \(M = \{\text{takes Finite Math}\}\) and \(H = \{\text{takes History}\}\). \(P(M) = 0.20\), \(P(H) = 0.30\), \(P(M \cap H) = 0.05\). Part a — Conditioning on History: \[P(M \mid H) = \frac{0.05}{0.30} = \frac{1}{6} \approx 0.167.\] Part b — Conditioning on Finite Math: \[P(H \mid M) = \frac{0.05}{0.20} = \frac{1}{4} = 0.25.\] Answer: a) \(1/6\); b) \(1/4\).Problem 8.4.9 Solution
Let \(A = \{\text{passes Accounting}\}\) and \(E = \{\text{passes English}\}\). \(P(A) = 0.60\), \(P(E) = 0.70\), \(P(A \cap E) = 0.30\). Part a: \[P(A \mid E) = \frac{0.30}{0.70} = \frac{3}{7} \approx 0.429.\] Part b: \[P(E \mid A) = \frac{0.30}{0.60} = \frac{1}{2} = 0.50.\] Answer: a) \(3/7\); b) \(1/2\).Problem 8.4.10 Solution
Step 1 — Use the multiplication rule: \[P(E \cap F) = P(F)\, P(E \mid F) = 0.4 \times 0.3 = 0.12.\] Answer: \(0.12\).Problem 8.4.11 Solution
Step 1 — Mutually exclusive means \(P(E \cap F) = 0\): \[P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{0.3} = 0.\] Answer: \(0\).Problem 8.4.12 Solution
\[P(F \mid E) = \frac{P(E \cap F)}{P(E)} = \frac{0.24}{0.6} = 0.4.\] Answer: \(0.4\).Problem 8.4.13 Solution
Step 1 — Solve the formula for \(P(F)\): \[P(F) = \frac{P(E \cap F)}{P(E \mid F)} = \frac{0.04}{0.1} = 0.4.\] Answer: \(0.4\).Problem 8.4.14 Solution
\(P(F) = 0.72\) (final exam), \(P(R) = 0.46\) (research paper), \(P(F \cap R) = 0.32\). \[P(F \mid R) = \frac{0.32}{0.46} = \frac{16}{23} \approx 0.696.\] Answer: \(16/23 \approx 0.696\).Problem 8.4.15 Solution
\[P(R \mid F) = \frac{0.32}{0.72} = \frac{4}{9} \approx 0.444.\] Answer: \(4/9 \approx 0.444\).Problem 8.4.16 Solution
Step 1 — Reduce the sample space to "first born is a boy": \(\{BBB, BBG, BGB, BGG\}\), so \(n = 4\). Step 2 — Count outcomes with exactly two boys: \(BBG\) and \(BGB\) (the third child is a girl). The interpretation "two boys" here means exactly two — i.e., two boys and one girl. \[P = \frac{2}{4} = \frac{1}{2}.\] Answer: \(1/2\).Problem 8.4.17 Solution
Step 1 — Condition on "at least one girl": all 8 outcomes except \(BBB\), so \(n = 7\). Step 2 — Count "all girls": only \(GGG\), so 1 outcome. \[P = \frac{1}{7}.\] Answer: \(1/7\).Problem 8.4.18 Solution
Step 1 — Condition on "first born is a boy": \(\{BBB, BBG, BGB, BGG\}\), \(n = 4\). Step 2 — Count "both sexes" within those: exclude \(BBB\) (all boys), leaving \(\{BBG, BGB, BGG\}\), so 3 outcomes. \[P = \frac{3}{4}.\] Answer: \(3/4\).Problem 8.4.19 Solution
Step 1 — Condition on "children of both sexes": exclude \(BBB\) and \(GGG\), giving 6 outcomes. Step 2 — Count "all boys" within those: \(BBB\) is excluded, so 0 outcomes. \[P = \frac{0}{6} = 0.\] Answer: \(0\) ("all boys" contradicts "both sexes").Problem 8.4.20 Solution
Step 1 — Condition on age 65+ (T row, total 528). Some-college count for that row: 84. \[P(C \mid T) = \frac{84}{528} = \frac{7}{44}.\] Answer: \(7/44 \approx 0.159\).Problem 8.4.21 Solution
Step 1 — Condition on Associate degree (A column, total 197). S row entry in that column: 80. \[P(S \mid A) = \frac{80}{197}.\] Answer: \(80/197 \approx 0.406\).Problem 8.4.22 Solution
Step 1 — Joint probability uses the grand total: \[P(C \cap T) = \frac{84}{2096} = \frac{21}{524}.\] Answer: \(21/524 \approx 0.040\).Problem 8.4.23 Solution
Step 1 — Condition on Bachelor (B column, total 418). R row entry: 188. \[P(R \mid B) = \frac{188}{418} = \frac{94}{209}.\] Answer: \(94/209 \approx 0.450\).Problem 8.4.24 Solution
Step 1 — Condition on age 25–44 (R row, total 796). Bachelor count: 188. \[P(B \mid R) = \frac{188}{796} = \frac{47}{199}.\] Answer: \(47/199 \approx 0.236\).Problem 8.4.25 Solution
Step 1 — Condition on age 45–64 (S row, total 772). Graduate count: 67. \[P(G \mid S) = \frac{67}{772}.\] Answer: \(67/772 \approx 0.087\).Problem 8.4.26 Solution